For `0 le x le (pi)/(2)` , the number of values of x satisfying the equation
tan x+ sec x = cos x is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( \tan x + \sec x = \cos x \) for \( 0 \leq x \leq \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
- \( \tan x = \frac{\sin x}{\cos x} \)
- \( \sec x = \frac{1}{\cos x} \)

Substituting these identities into the equation gives us:
\[ \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \cos x \]

### Step 2: Combine the left side
The left side can be combined since they have a common denominator:
\[ \frac{\sin x + 1}{\cos x} = \cos x \]

### Step 3: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ \sin x + 1 = \cos^2 x \]

### Step 4: Use the Pythagorean identity
We know from the Pythagorean identity that:
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting this into our equation gives:
\[ \sin x + 1 = 1 - \sin^2 x \]

### Step 5: Rearrange the equation
Rearranging the equation results in:
\[ \sin^2 x + \sin x = 0 \]

### Step 6: Factor the equation
Factoring out \(\sin x\) gives:
\[ \sin x (\sin x + 1) = 0 \]

### Step 7: Solve for \(\sin x\)
Setting each factor to zero gives us:
1. \( \sin x = 0 \)
2. \( \sin x + 1 = 0 \) (which gives \( \sin x = -1 \), not possible in the interval \( [0, \frac{\pi}{2}] \))

### Step 8: Find the value of \(x\)
The only solution from \( \sin x = 0 \) in the interval \( [0, \frac{\pi}{2}] \) is:
\[ x = 0 \]

### Conclusion
Thus, the number of values of \(x\) satisfying the equation \( \tan x + \sec x = \cos x \) in the interval \( [0, \frac{\pi}{2}] \) is **1** (specifically \(x = 0\)).

---